### Two Examples of Borel Partially Ordered Sets with the Countable Chain Condition

Stevo Todorcevic
1991 Proceedings of the American Mathematical Society
We define an open symmetric two-place relation on the reals such that the reals cannot be covered by countably many sets of related elements, but there is no uncountable set of unrelated elements. The poset & of finite sets of related elements satisfies the countable chain condition but it may fail to have the property K, i.e., a substantial irregularity can be injected in 5a . We construct two examples of Borel posets that satisfy the countable chain condition but violate certain natural
more » ... thenings of this condition. The first example, discovered by the author several years ago to answer a question of L. Harrington and S. Shelah, is a Borel version of the poset presented in [5, 9.10]. Let 7rQ be the set of all subsets of the rationals ordered by x < y iff there is a q in y such that x = {p £ y : p < q} . Let 3°^ be the set of all finite antichains of the pseudotree nQ>. The ordering of S°0 is the inclusion. Clearly, ^ is Borel. Claim 1. ^0 is a ccc poset. Proof. Almost identical to the standard proof for the poset of all finite antichains of an Aronszajn tree. (See [5, 9.10(iii)].) Claim 2. A3°q is not o-linked. Proof. Assume {JA^} is a sequence of linked subsets of £P0. Let qQ be any rational, say q0 = 0. If there is {t} in AjA0 such that sup t < q0, pick one and call it t0 ; otherwise t0 = {-1}. Pick a rational qx such that sup t0 < qx < q0. If there is {t} in 5AX such that t0 < t and sup/ < qx , pick one and call it tx, ... , and so on. Let iM = |J(. tt. Then {t^} £ Jz^ for all i. The second example is a Borel version of the Galvin-Hajnal poset as presented in [2]. A similar modification with a quite different motivation is given in [1]. The basic properties of the poset, however, are based on a completely different set of ideas ([3, §4], [4, §2]). Let 5A be the set of all converging sequences of